Another aspect that I find people tend to "forget" with costs of desalination related to power consumption, is the fact that this water has to be pumped up-hill. This is an unavoidable "cost", there is no way around this if you plan to use desalination, and desalination operation are almost invariable located at sea level. Forget about the immense amount of power required to generate the pressure to push the saline water through the membranes.
Let's take a look at the case of Los Angeles as a possible example. This is a very simple back of the envelope calculation considering that the whole of Los Angeles water supply would to be done through desalinated water, now I know that won't be the case, normally this additional sort of infrastructure is supplementary, remember this is to illustrate a point.
Using the data provided by Los Angeles Department of Power and Water (LADWP) it sold 193 billions of gallons of water to it's consumer in the 2009-2010 fiscal year. Taking the elevation of Los Angeles at 71 m (233 ft) location of the City Hall, as an approximate average for this calculation, calculating just the potential energy in joules required to simply elevate vertically this water, not considering any horizontal transportation or mechanical inefficiencies of pumps, etc... it requires 1.39 TJoules, this corresponds to 16.2 Mega Watts per day. Additionally this does not even consider providing any extra head or mains pressure...
Have in mind additionally that Los Angeles supplies itself with 70% non renewable power sources (http://www.ladwp.com/ladwp/cms/ladwp000509.jsp) which are only carbon emission producing.
Now this may seem a lot, or maybe not very much, if one considers the cost of which a KWatt/h is sold at. My idea is not evaluate if this is expensive or not, it is to illustrate that there is energy to be spent on something that can be avoided. As a percentage of the total power consumption of Los Angeles it might seem insignificant, however remember this is a pure mathematical physics model.
My simple calculations can be seen here:
https://spreadsheets.google.com/ccc?key=0ArEwIvQBUPQzdHROOEcyRUh4cUtiOWNHeHR2Q1lqZFE&hl=en_GB&authkey=CMz5oJMP
No matter how energy efficient you become, there is no way to avoid the energy cost of simple gravity.
Let's take a look at the case of Los Angeles as a possible example. This is a very simple back of the envelope calculation considering that the whole of Los Angeles water supply would to be done through desalinated water, now I know that won't be the case, normally this additional sort of infrastructure is supplementary, remember this is to illustrate a point.
Using the data provided by Los Angeles Department of Power and Water (LADWP) it sold 193 billions of gallons of water to it's consumer in the 2009-2010 fiscal year. Taking the elevation of Los Angeles at 71 m (233 ft) location of the City Hall, as an approximate average for this calculation, calculating just the potential energy in joules required to simply elevate vertically this water, not considering any horizontal transportation or mechanical inefficiencies of pumps, etc... it requires 1.39 TJoules, this corresponds to 16.2 Mega Watts per day. Additionally this does not even consider providing any extra head or mains pressure...
Have in mind additionally that Los Angeles supplies itself with 70% non renewable power sources (http://www.ladwp.com/ladwp/cms/ladwp000509.jsp) which are only carbon emission producing.
Now this may seem a lot, or maybe not very much, if one considers the cost of which a KWatt/h is sold at. My idea is not evaluate if this is expensive or not, it is to illustrate that there is energy to be spent on something that can be avoided. As a percentage of the total power consumption of Los Angeles it might seem insignificant, however remember this is a pure mathematical physics model.
My simple calculations can be seen here:
https://spreadsheets.google.com/ccc?key=0ArEwIvQBUPQzdHROOEcyRUh4cUtiOWNHeHR2Q1lqZFE&hl=en_GB&authkey=CMz5oJMP
No matter how energy efficient you become, there is no way to avoid the energy cost of simple gravity.
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